Answer to Question #172454 in Mechanical Engineering for Diocel Ann Heria

Question

7. A stone is thrown with an initial velocity of

100 fps upward at 60 to the horizontal. Compute

the radius of curvature of its path at the point

where it is 50 ft horizontally from its initial

position.

Answer 1

We can use the equations of motion for constant acceleration if we split the initial velocity into its horizontal and vertical components.

u_x = u*cosθ

100 fps = 30.48 m/s

u_x = 30.48 m/s * cos 60

u_x = 15.24 m/s

u_y = u*sinθ

u_y = 30.48 m/s * sin60

u_y = 26.40 m/s

In the horizontal direction, assuming no air resistance, there will be no resultant force. This means that the acceleration will be zero in the horizontal direction, so we can just use:

s = ut + at²/2

s_x = u_xt + 0t²/2

s_x = u_xt

50 ft = 15.24 m

15.24 m = 15.24 m/s * t

t = 1 s

The time of flight of the stone is therefore 10 seconds. We can now use the same equation but for the vertical component. In the vertical component, we can use g for acceleration.

s_y = u_yt + gt²/2

s_y = 26.40 m/s * 1 s - (9.8 m/s² * (1 s)²)/2

s_y = 21.5 m

r = 21.5 m